```PHY 206 Spring 2003 HW #8 Solutions
19. **A 1500-kg car moves west at 20 m/s and a 3000-kg truck moves east at 16 m/s. What
is the velocity of the center of mass of the system.
The center of mass velocity is given by
Mvcm  m1v1  m2v2   3000kg 16m/s   1500kg  20m/s  . Solving we find vcm  4 ms .
23.
In the Atwood’s machine in Fig. 8-51 we introduce the mass of the pulley, mc. (a) Find
the acceleration of the center of mass of the two-block plus pulley system. (b) Find the force exerted
by the support. (c) Find the tension in the connecting string
For the acceleration of the center of mass, we have Macm  m1a1  m2a2  mcac . Since the pulley is
stationary, ac = 0. From Chap. 4, we know that the acceleration of the two blocks is given by
a1  a2 
m1  m2
g . The acceleration of the center of mass is thus
m1  m2
 m1  m2 
 m  m2   m1  m2 
 1
g . The force exerted by the support is

g 
 m1  m2  m1  m2  mc 
 M   m1  m2 
2
acm
 4m1m2 
 g . The
 m1  m2 
2m1m2
g.
tension in the string is independent of the pulley, so we use the result from Chap. 4, T 
m1  m2
found from Fnet  Mg  F  Macm , which leads to F  M  g  acm   mc g  
Thus F = mcg + 2T, using the result for F above.
25.
(a) True
(b) True
(c) True
27.
A man on frictionless skates needs to get to the edge. How can he do so?
Take off anything he is wearing and throw it in the opposite direction he wants to go.
35. **Launched out of a cannon at 24 m/s and an angle of 30o, Marcello (70 kg) collides at the top of
his trajectory with Tina (50kg) and then lands in a net at the same height as the cannon. How far away
do they land?
The collision is inelastic (they stick together). Since the collision takes place at the top of the
trajectory, there is only horizontal-direction motion to consider in the collision. We use
mM vMxi   mT  mM  vx where vMxi = 24 cos 30o m/s = 20.8 m/s. Solving, we find vx = 12.1 m/s.
Returning to two-dimensional motion, the time it takes for Marcello to reach the highest point is
t
v  v0
 1.22 s , where a = -9.81m/s. Thus he travels, in the x-direction, x1  vMxit  25.4m . Since
a
the masses don’t matter, it will also take 1.22 s to reach the net, and in that time the pair travels
x2  vxt  14.8m . The total distance traveled is xT = x1 + x2 = 40.2 m.
48. **When a 0.15 kg baseball is hit, its velocity changes from 20 m/s to -20 m/s. The time of contact
might be 1.3 ms. What is the impulse delivered, and what is the average force?
kg-m
Impulse is defined as I  p , so in this case we have I   0.15kg   20 ms    0.15kg   20 ms   6 s .
The average force is Fav 
49.
p
 4600 N .
t
A 300-g handball moving with a speed of 5.0 m/s strikes the wall at an angle of 40 and then
bounces off with the same speed at the same angle. It is in contact with the wall for 2 ms. What is the
average force exerted by the ball on the wall?
The perpendicular component of the momentum is important here, pperp = mv cos = 1.15 kg-m/s. In bouncing
off, the change in momentum is given by p = pf – pi = (-1.15 kg-m/s) – (1.15 kg-m/s) = -2.3 kg-m/s. The
magnitude of the average force is p/t = 1150 N.
55.
(a) False
(b) True
(c) True
67.
**A 3-kg block with speed 4 m/s makes an elastic collision with a stationary block of mass 2 kg.
Use conservation of momentum and the relative velocity relations to find the velocity of each block after the
collision. Check by using conservation of kinetic energy.
The relative speed of approach is given by vapproach = (0 – 4) m/s = -vrecession = -v2f + v1f. Conservation of
momentum is written m1v1i  m1v1 f  m2v2 f . Solving these two together, we find v2f = 4.8 m/s, and
therefore v1f = 0.8 m/s. For this special case of one mass initially at rest, equations 8-30 are relevant; be
careful about using this result, since it is valid only for the special case. The initial kinetic energy is given by
Ki  12 m1v12i  24J . The final kinetic energy is K f  12 m1v12f  12 m2v22 f  0.96J  23.04J=24 J .
69.
A block of mass m1 = 2 kg has a speed of 10 m/s. Moving in the same direction with a speed of
3 m/s is a block of mass m2 = 5 kg. A massless spring, k = 1120 N/m is attached to the second block. (a)
Before the collision, what is the velocity of the center of mass? (b) After the collision, the spring is
compressed by a maximum amount x. What is x? (c) The blocks separate. What are the final velocities of
the blocks?
Initially the velocity of the center of mass is vcm 
m1v1  m2v2
 5m/s . At the point of maximum spring
M
compression, the two blocks move together with the same velocity, which must necessarily be the center-ofmass velocity. Conservation of energy tells us that Ki  ET  K cm  U spring . The initial kinetic energy is Ki =
½ m1v1i2 +½ m2v2i2 = 100 J + 22.5 J = 122.5 J. The center-of-mass kinetic energy is Kcm = ½ Mvcm2 = 87.5 J.
Thus the spring potential energy is Usp = 35 J = ½ k (x)2. Using this relation, x = 0.25 m. The total kinetic
energy of the center of mass remains constant, as does the momentum. The relative speed of approach is 3 m/s
– 10 m/s = -7 m/s. This is equal to -vrecession = -(v2f – v1f). From conservation of momentum,
Mvcm  m1v1 f  m2v2 f . Substituting v2 f  7m/s  v1 f , we solve to find v1f = 0 m/s and thus v2f = 7 m/s.
105.
**A 4-kg fish swims at 1.5 m/s to the right and swallows a 1.2 kg fish swimming at 3 m/s to
the left. What is the velocity of the larger fish after lunch?
This is an inelastic collision, so we know that the final velocity is common to the two fish.
Momentum is conserved, m1v1i  m2v2i   m1  m2  v f . Taking into the account the velocity sign for
fish 2, we find vf = 0.46 m/s.
108.
A 3000-kg shark moves at 3.0 m/s vertically downward and eats a 200-kg fish moving at 8.0
m/s horizontally. What is the angle and final speed at which the shark moves after dinner?
This is an inelastic collision in two dimensions. We can write m2v2 x   m1  m2  vxf , since
the shark (fish 1) has no initial x-direction motion. Solving gives us vxf = 0.5 m/s. In the y-direction
we have m1v1 y   m1  m2  vyf , which gives us vyf = -2.81 m/s. The angle with respect to the vertical
2
 0.5 
2
m
  10.1 and the speed is v f  0.5   2.81  2.84 s .
2.81


is   tan 1 
112.
**A pendulum with bob of mass 1 kg and length 2 m swings from a horizontal position and hits a
block of mass 1 kg resting on a rough shelf ( = 0.1). Assuming the collision is perfectly elastic, (a) what is
the velocity of the block just after impact? (b) How far does the block move before coming to rest?
The initial potential energy of the pendulum is U = mgh = 19.6 J. As the pendulum reaches the lowest point, it
has kinetic energy equal to this potential energy. We can find the speed from conservation of energy:
v1  2 gh  6.26m/s . (a) From conservation of momentum we have Pi = Pf. Using Eqns. 8-30, we see that
v1f = 0 m/s and v2f = 6.26 m/s. (b) The initial kinetic energy of the block is thus Ki = 19.6 J. This is the energy
dissipated by friction, -fs = -19.6 J. The frictional force is f = Fn = mg. We can solve this for the distance
s = 20 m.
114.
**A 15-g bullet traveling at 500 m/s hits an 0.8-kg block balanced on the edge of a table of
height h = 0.8 m. How far does the block go before hitting the floor?
In the inelastic collision we have mb vi   mb  mB  v f from which we find vf = 9.2 m/s. The
height above the floor is 0.8 m so we can calculate the time needed to hit the floor from y 
which allows us to find t = 0.40 s. The x-direction distance is found from
x  vxt   9.2 ms   0.4s   3.72m .
1 2
a yt ,
2
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